I just want to see if my work is correct.
This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.
4.8 #1
y`` + y = 0
a.
e^xsolve with: e^x
e^x y`` + y = 0
a y"+b y' +c y=0
a y’’ +b y’ + c =0
a e^(y’’ x) + b e^(y’ x)
e^(y)(a cos(x) +b sin(x))
This is not a solution
b.Sin x
sin x y'' + y = 0
1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
y(x)+sin(x) = cos(x) y''(x)
Solution
c.Cos x
cos x y'' + y = 0
1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution
d.Sin x – cos x
-1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0
Solution
[QUOTE=pikachumanson;533340]I just want to see if my work is correct.
This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.
4.8 #1
y`` + y = 0
What does that mean?Is this the differential equation you want to solve?
[quote[ a.
e^xsolve with: e^x
Just put in front of y''? Why?e^x y`` + y = 0
Now you have taken the exponential of each derivative. Again, why? And where did a, b, and c suddenly come from?a y"+b y' +c y=0
a y’’ +b y’ + c =0
a e^(y’’ x) + b e^(y’ x)
I agree with mr. fantastic- what you wrote makes no sense at all!e^(y)(a cos(x) +b sin(x))
This is not a solution
b.Sin x
sin x y'' + y = 0
1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
y(x)+sin(x) = cos(x) y''(x)
Solution
c.Cos x
cos x y'' + y = 0
1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution
d.Sin x – cos x
-1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0
Solution